(A) The oxidation state of $Ni$ in $[NiCl_4]^{2-}$ is $+2$. The electronic configuration of $Ni^{2+}$ is $[Ar] 3d^8 4s^0$. As $Cl^{-}$ is a weak field

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(A) The oxidation state of $Ni$ in $[NiCl_4]^{2-}$ is $+2$. The electronic configuration of $Ni^{2+}$ is $[Ar] 3d^8 4s^0$. As $Cl^{-}$ is a weak field ligand,pairing of electrons will not occur. The $Ni^{2+}$ ion in the complex undergoes $sp^3$ hybridisation using one $4s$ and three $4p$ orbitals. Thus,the geometry is tetrahedral. Since there are $2$ unpaired electrons in the $3d$ orbitals,the complex is paramagnetic.

Class 12 Chemistry Coordination Compounds Hybridisation and Geometry

Medium

The geometry and magnetic property of $[NiCl_4]^{2-}$,respectively are

KVPY 2017 All KVPY

A
tetrahedral,paramagnetic
B
tetrahedral,diamagnetic
C
square planar,paramagnetic
D
square planar,diamagnetic

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Class 12

Chemistry Questions

Chapter

Coordination Compounds

Subtopic

Hybridisation and Geometry

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Which of the following is not a square planar complex?

Easy

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For $K_3[Co(CN)_6]$,which of the following statements is not correct?

Medium

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Which of the following has a square planar geometry? $(Co = 27, Ni = 28, Fe = 26, Pt = 78)$

Medium

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Match the coordination number and type of hybridisation with the distribution of hybrid orbitals in space based on Valence Bond Theory.

Coordination number and type of hybridisation Distribution of hybrid orbitals in space

$a. 4, sp^3$ $i. \text{trigonal bipyramidal}$

$b. 4, dsp^2$ $ii. \text{octahedral}$

$c. 5, sp^3d$ $iii. \text{tetrahedral}$

$d. 6, d^2sp^3$ $iv. \text{square planar}$

Select the correct option.

Coordination number and type of hybridisation	Distribution of hybrid orbitals in space
$a. 4, sp^3$	$i. \text{trigonal bipyramidal}$
$b. 4, dsp^2$	$ii. \text{octahedral}$
$c. 5, sp^3d$	$iii. \text{tetrahedral}$
$d. 6, d^2sp^3$	$iv. \text{square planar}$